Cfrgtkky

I want to solve the following differential equation with initial conditions:

$$frac{mathrm{d}^2 y}{mathrm{d} x^2}=frac{x , y(x)}{sqrt{1-x}}$$
But do not know how to actually solve it. Any suggestion?

One thing you could do is look for a power series solution. This is a linear ODE with an irregular singularity at $x=1$, but if we look for a solution centered at $x=0$, so $|x|<1$, then we should be okay. Note this makes sense to do given where $frac{x}{sqrt{1-x}}$ is defined:

Suppose $y = sum_na_nx^nRightarrow y'' = sum_n n(n-1)a_n x^{n-2}$

Also, $frac{x}{sqrt{1-x}} = sum_{k=0}^infty {-frac{1}{2} choose k} x^{k+1}$, by the binomial series.

Now we can multipy these expressions and solve for the coefficients $a_n$:

$y'' =displaystyle sum_n n(n-1)a_n x^{n-2}=left(displaystylesum_{k=0}^infty {-frac{1}{2} choose k} x^{k+1}right) left(displaystylesum_na_nx^nright)$.

It is messy, but should work out with a bit of computation. Hopefully , it suffices for your needs.

Hint:

Let $u=sqrt{1-x}$ ,

Then $dfrac{dy}{dx}=dfrac{dy}{du}dfrac{du}{dx}=-dfrac{1}{2sqrt{1-x}}dfrac{dy}{du}$

$dfrac{d^2y}{dx^2}=dfrac{d}{dx}left(-dfrac{1}{2sqrt{1-x}}dfrac{dy}{du}right)=-dfrac{1}{2sqrt{1-x}}dfrac{d}{dx}left(dfrac{dy}{du}right)+dfrac{1}{4(1-x)^frac{3}{2}}dfrac{dy}{du}=-dfrac{1}{2sqrt{1-x}}dfrac{d}{du}left(dfrac{dy}{du}right)dfrac{du}{dx}+dfrac{1}{4(1-x)^frac{3}{2}}dfrac{dy}{du}=-dfrac{1}{2sqrt{1-x}}dfrac{d^2y}{du^2}left(-dfrac{1}{2sqrt{1-x}}right)+dfrac{1}{4(1-x)^frac{3}{2}}dfrac{dy}{du}=dfrac{1}{4(1-x)}dfrac{d^2y}{du^2}+dfrac{1}{4(1-x)^frac{3}{2}}dfrac{dy}{du}$

$thereforedfrac{1}{4(1-x)}dfrac{d^2y}{du^2}+dfrac{1}{4(1-x)^frac{3}{2}}dfrac{dy}{du}=dfrac{xy}{sqrt{1-x}}$

$dfrac{1}{4u^2}dfrac{d^2y}{du^2}+dfrac{1}{4u^3}dfrac{dy}{du}-dfrac{(1-u^2)y}{u}=0$

$udfrac{d^2y}{du^2}+dfrac{dy}{du}+4u^2(u^2-1)y=0$

Consider generally change the abscissa and the ordinate:

Let $u=f(v)$ ,

Then $dfrac{dy}{du}=dfrac{dfrac{dy}{dv}}{dfrac{du}{dv}}=dfrac{1}{f'(v)}dfrac{dy}{dv}$

$dfrac{d^2y}{du^2}=dfrac{d}{du}left(dfrac{1}{f'(v)}dfrac{dy}{dv}right)=dfrac{dfrac{d}{dv}left(dfrac{1}{f'(v)}dfrac{dy}{dv}right)}{dfrac{du}{dv}}=dfrac{dfrac{1}{f'(v)}dfrac{d^2y}{dv^2}-dfrac{f''(v)}{(f'(v))^2}dfrac{dy}{dv}}{f'(v)}=dfrac{1}{(f'(v))^2}dfrac{d^2y}{dv^2}-dfrac{f''(v)}{(f'(v))^3}dfrac{dy}{dv}$

$thereforedfrac{f(v)}{(f'(v))^2}dfrac{d^2y}{dv^2}-dfrac{f(v)f''(v)}{(f'(v))^3}dfrac{dy}{dv}+dfrac{1}{f'(v)}dfrac{dy}{dv}+4(f(v))^2((f(v))^2-1)y=0$

$dfrac{d^2y}{dv^2}+left(dfrac{f'(v)}{f(v)}-dfrac{f''(v)}{f'(v)}right)dfrac{dy}{dv}+4f(v)((f(v))^2-1)(f'(v))^2y=0$

For example when take $f(v)=dfrac{pv+q}{rv+s}$ , the ODE becomes

$dfrac{d^2y}{dv^2}+left(dfrac{ps-qr}{(pv+q)(rv+s)}+dfrac{2r}{rv+s}right)dfrac{dy}{dv}+dfrac{4(ps-qr)^2(pv+q)((pv+q)^2-(rv+s)^2)}{(rv+s)^7}y=0$